Worked First-Passage and Interval-Exit Problems

A Calculation Framework for First Passage and Exit

First-passage and interval-exit questions often ask for related but distinct quantities: the probability of ever reaching a level, a discounted transform of the time to reach it, the probability of leaving a bounded interval at a specified endpoint, or the mean time until either endpoint is reached. The calculations below use explicit formulas and stopped Ito arguments to keep these distinctions visible.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider $X_t=x+\mu t+\sigma B_t$, where $x,\mu\in\mathbb{R}$ and $\sigma>0$. Whenever an Ito integral $\int_0^t H_s\,dB_s$ appears, it is the Ito integral of a progressively measurable adapted integrand in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^t H_s^2\,ds<\infty$. The drift coefficient is $\mu$; the diffusion coefficient is $\sigma$.

Definition: For an upper level $a>x$, let $\tau_a=\inf\{t\ge0:X_t=a\}$, with $\inf\varnothing=\infty$. For an interval $l<x<u$, let $\tau=\inf\{t\ge0:X_t\notin(l,u)\}$. Because $X$ has continuous paths, a finite interval exit occurs at $l$ or $u$.

One-Sided Hitting: Probability and Transform

The transform $\mathbb{E}[e^{-q\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]$, for $q\ge0$, combines the probability of ever hitting $a$ with information about the hitting time. At $q=0$, the exponential factor is one on the hitting event, so the transform is exactly the eventual hitting probability.

Theorem: If $a>x$, $\sigma>0$, and $q\ge0$, then $$ \mathbb{E}\!\left[e^{-q\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}\right] =\exp\!\left(\frac{\mu-\sqrt{\mu^2+2q\sigma^2}}{\sigma^2}(a-x)\right). $$ In particular, the probability of ever hitting $a$ is $1$ when $\mu\ge0$, and is $\exp(2\mu(a-x)/\sigma^2)$ when $\mu<0$.

Proof. Set $r=(-\mu+\sqrt{\mu^2+2q\sigma^2})/\sigma^2$, so $r\ge0$, and define $g(y)=e^{r(y-a)}$. Its derivatives are $g'(y)=rg(y)$ and $g''(y)=r^2g(y)$. The choice of $r$ gives $\mu r+\tfrac12\sigma^2r^2=q$. Ito's formula, including its quadratic-variation correction, therefore gives $$ dg(X_t)=g'(X_t)\sigma\,dB_t+ \left(\mu g'(X_t)+\tfrac12\sigma^2g''(X_t)\right)dt =\sigma r g(X_t)\,dB_t+qg(X_t)\,dt. $$ Applying the product rule to $e^{-qt}g(X_t)$, the $dt$ terms cancel. Up to $\tau_a$, $X_t\le a$, so $0<g(X_{t\wedge\tau_a})\le1$. On each finite time interval the stopped stochastic-integral integrand $e^{-qs}\sigma r g(X_s)\mathbf{1}_{\{s\le\tau_a\}}$ is bounded and hence in $\mathcal{L}^2$. The Ito integral is a square-integrable martingale: its increments have conditional expectation zero by the Ito integral construction, and its second moment is finite by the Ito isometry. Consequently, for every $t\ge0$, $$ g(x)=\mathbb{E}\!\left[e^{-q(t\wedge\tau_a)}g(X_{t\wedge\tau_a})\right] =\mathbb{E}\!\left[e^{-q\tau_a}\mathbf{1}_{\{\tau_a\le t\}}\right] +\mathbb{E}\!\left[e^{-qt}g(X_t)\mathbf{1}_{\{\tau_a>t\}}\right]. $$ For $q>0$, the second term is at most $e^{-qt}$ and tends to zero. For $q=0$, if $\mu>0$, then $X_t\to+\infty$ almost surely; if $\mu=0$, then $\limsup_{t\to\infty}X_t=+\infty$ almost surely. In either case continuity implies $\tau_a<\infty$ almost surely, so the second term is at most $\mathbb{P}(\tau_a>t)\to0$. If $q=0$ and $\mu<0$, then $r=-2\mu/\sigma^2>0$ and $X_t\to-\infty$ almost surely. On $\{\tau_a>t\}$, $g(X_t)\to0$; since $0\le g(X_t)\le1$ before hitting, dominated convergence again makes the second term tend to zero. Thus the transform is $g(x)$. Rewriting $g(x)=e^{-r(a-x)}$ gives the displayed formula. At $q=0$, $r=0$ for $\mu\ge0$, while $r=-2\mu/\sigma^2$ for $\mu<0$, yielding the stated hitting probabilities. $\square$

Worked Example: A Negative-Drift Hitting Transform

Let $X_t=1-t+2B_t$, and let $a=3$. Find the transform at $q=3$ and the probability of ever hitting $3$.

Here $x=1$, $\mu=-1$, $\sigma^2=4$, and $a-x=2$. For $q=3$, the square root in the theorem is $$ \sqrt{\mu^2+2q\sigma^2}=\sqrt{(-1)^2+2\cdot3\cdot4}=\sqrt{25}=5. $$ Thus the exponent is $\frac{-1-5}{4}\cdot2=-3$, and the transform is $e^{-3}$.

At $q=0$, $\mu<0$, so the hitting probability is $$ \exp\!\left(\frac{2(-1)(2)}{4}\right)=e^{-1}. $$ Check: the transform at positive $q$ is $e^{-3}<e^{-1}$, as expected because discounting can only reduce the expectation. Also the characteristic root for $q=3$ is $r=(1+5)/4=3/2$; $e^{-r(a-x)}=e^{-(3/2)2}=e^{-3}$, agreeing with the calculation.

Exit Side from a Bounded Interval

On a bounded interval, the process exits in finite time almost surely. One direct justification is useful for the stopped calculations that follow. Fix any time step $h>0$. From any point in $[l,u]$, an increment $\mu h+\sigma(B_{s+h}-B_s)>u-l$ forces the endpoint at time $s+h$ to lie above $u$, and continuity then forces an exit during that step. This event has a probability $p>0$ independent of the starting point and of the past, since the increment is normal with variance $\sigma^2h>0$. Independent increments give $\mathbb{P}(\tau>nh)\le(1-p)^n$, so $\tau<\infty$ almost surely.

Theorem: For $l<x<u$, the probability of exiting at the upper endpoint is $$ \mathbb{P}(X_\tau=u)= \begin{cases} \dfrac{1-e^{-2\mu(x-l)/\sigma^2}}{1-e^{-2\mu(u-l)/\sigma^2}},&\mu\ne0,\\[1.2ex] \dfrac{x-l}{u-l},&\mu=0. \end{cases} $$

Proof. Write $L=u-l$ and $y=x-l$. For $\mu\ne0$, define $h(z)=(1-e^{-2\mu(z-l)/\sigma^2})/(1-e^{-2\mu L/\sigma^2})$ on $[l,u]$. The denominator is nonzero because $\mu L\ne0$. This function has $h(l)=0$, $h(u)=1$, and direct differentiation gives $\mu h'(z)+\tfrac12\sigma^2h''(z)=0$. For $\mu=0$, use $h(z)=(z-l)/L$, which has the same boundary values and satisfies $\tfrac12\sigma^2h''(z)=0$. Ito's formula for $h(X_{t\wedge\tau})$ has drift term $(\mu h'+\tfrac12\sigma^2h'')dt=0$. Its stochastic-integral integrand is bounded on $[l,u]$, so on each finite time interval it is in $\mathcal{L}^2$, and the integral is a martingale with conditional mean-zero increments. Hence $\mathbb{E}[h(X_{t\wedge\tau})]=h(x)$. Since $h$ is bounded, $\tau<\infty$ almost surely, and $X_\tau\in\{l,u\}$, dominated convergence gives $\mathbb{E}[h(X_\tau)]=h(x)$. The boundary values imply $h(X_\tau)=\mathbf{1}_{\{X_\tau=u\}}$. Substitution of $z=x$ gives the formula. $\square$

Worked Example: Exit Side and Mean Time with Positive Drift

Take $X_t=1+t+\sqrt{2}B_t$ in the interval $(0,2)$. Compute the probability of upper exit and the mean exit time.

For the exit probability, $\mu=1$, $\sigma^2=2$, $l=0$, $u=2$, and $x=1$. Therefore $$ \mathbb{P}(X_\tau=2) =\frac{1-e^{-2\cdot1\cdot1/2}}{1-e^{-2\cdot1\cdot2/2}} =\frac{1-e^{-1}}{1-e^{-2}} =\frac{1}{1+e^{-1}}. $$ The last equality follows from $1-e^{-2}=(1-e^{-1})(1+e^{-1})$. The lower-exit probability is $1-\mathbb{P}(X_\tau=2)=e^{-1}/(1+e^{-1})$; the two probabilities sum to one.

For the mean exit time, the formula proved below gives $$ \mathbb{E}[\tau] =\frac{2}{1}\frac{1-e^{-1}}{1-e^{-2}}-\frac{1}{1} =\frac{2}{1+e^{-1}}-1 =\frac{1-e^{-1}}{1+e^{-1}}. $$ Check: this is positive because $0<e^{-1}<1$. It is less than $1$, since $1-e^{-1}<1+e^{-1}$. These checks are consistent with a positive finite mean exit time.

Mean Exit Time

Theorem: Let $L=u-l$ and $y=x-l$, where $l<x<u$. For $X_t=x+\mu t+\sigma B_t$, $\sigma>0$, the mean time to exit $(l,u)$ is $$ \mathbb{E}[\tau]= \begin{cases} \dfrac{L}{\mu}\dfrac{1-e^{-2\mu y/\sigma^2}}{1-e^{-2\mu L/\sigma^2}}-\dfrac{y}{\mu},&\mu\ne0,\\[1.2ex] \dfrac{y(L-y)}{\sigma^2},&\mu=0. \end{cases} $$

Proof. Define $m(z)$ on $[l,u]$ by the right-hand side with $y$ replaced by $z-l$. It is twice continuously differentiable, has $m(l)=m(u)=0$, and satisfies $$ \mu m'(z)+\tfrac12\sigma^2m''(z)=-1. $$ For $\mu\ne0$, this follows by differentiating the exponential expression: with $k=2\mu/\sigma^2$ and $A=L/(\mu(1-e^{-kL}))$, write $m(z)=A(1-e^{-k(z-l)})-(z-l)/\mu$. Then $m'=Ak e^{-k(z-l)}-1/\mu$ and $m''=-Ak^2e^{-k(z-l)}$. Consequently the exponential terms in $\mu m'+\tfrac12\sigma^2m''$ cancel because $\mu k=\tfrac12\sigma^2k^2$, leaving $-1$. For $\mu=0$, $m(z)=(z-l)(u-z)/\sigma^2$, so $m''(z)=-2/\sigma^2$ and $\tfrac12\sigma^2m''(z)=-1$.

Apply Ito's formula to $m(X_{t\wedge\tau})$. Its stochastic-integral integrand is $\sigma m'(X_s)\mathbf{1}_{\{s\le\tau\}}$, bounded because $m'$ is continuous on the compact interval $[l,u]$. On each finite time interval it is in $\mathcal{L}^2$, so the Ito integral has expectation zero. The drift calculation gives $$ \mathbb{E}[m(X_{t\wedge\tau})]=m(x)-\mathbb{E}[t\wedge\tau]. $$ As $t\to\infty$, $X_{t\wedge\tau}\to X_\tau$ almost surely and $m(X_\tau)=0$. Boundedness of $m$ gives $\mathbb{E}[m(X_{t\wedge\tau})]\to0$ by dominated convergence. Also $t\wedge\tau\uparrow\tau$, so monotone convergence gives $\mathbb{E}[t\wedge\tau]\to\mathbb{E}[\tau]$. Thus $\mathbb{E}[\tau]=m(x)$, as claimed. $\square$

Worked Example: Driftless Exit from a Symmetric Interval

Let $X_t=1+2B_t$, and let the interval be $(-2,4)$. Find the probability of exiting at $4$ and the mean exit time.

Here $L=4-(-2)=6$, $y=x-l=1-(-2)=3$, and $\sigma^2=4$. Since $\mu=0$, the upper-exit probability is $y/L=3/6=1/2$, and the mean exit time is $$ \mathbb{E}[\tau]=\frac{y(L-y)}{\sigma^2} =\frac{3(6-3)}{4} =\frac{9}{4}. $$ Check: the interval is symmetric about $1$, so driftless Brownian motion starting at its midpoint has equal probabilities of exiting either side. The mean-time function is $m(z)=(z+2)(4-z)/4$; at $z=1$ this gives $3\cdot3/4=9/4$, agreeing with the formula.

A Common Optional-Stopping Pitfall

A martingale identity at a fixed time cannot automatically be evaluated at an unbounded hitting time. For a concrete example, let $X_t=t+B_t$, start at $0$, and let $\tau_1$ be the first time $X$ reaches $1$. The process $M_t=e^{-2X_t}$ is a martingale: Ito's formula gives $dM_t=-2M_t\,dB_t$, with zero drift. A tempting but invalid calculation is $\mathbb{E}[M_{\tau_1}]=M_0=1$, followed by $M_{\tau_1}=e^{-2}$, which would claim $e^{-2}=1$. The error is applying optional stopping without a valid boundedness or uniform-integrability argument at $\tau_1$.

The correct hitting-probability calculation uses the one-sided theorem: here $\mu=1>0$, so $\mathbb{P}(\tau_1<\infty)=1$. In the stopped finite-time identity, the term on $\{\tau_1>t\}$ is not discarded by a fixed-time martingale identity; its limiting behavior must be justified separately. The proof above does this explicitly, including the special zero-drift case where Brownian motion does not converge to $+\infty$, but has unbounded limsup.

Question	Quantity	Key condition or distinction
Ever hit an upper level?	$\mathbb{P}(\tau_a<\infty)$	Drift sign determines whether the probability is below one
Discounted first passage?	$\mathbb{E}[e^{-q\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]$	$q\ge0$; at $q=0$ this is the hitting probability
Which interval endpoint?	$\mathbb{P}(X_\tau=u)$	Use the driftless limit separately when $\mu=0$
How long until exit?	$\mathbb{E}[\tau]$	Finite for a bounded interval with $\sigma>0$

Check Your Understanding

In each question, the starting point lies strictly inside the stated interval or below the stated upper level.

For $X_t=x+\mu t+\sigma B_t$ with $\mu<0$, state the probability of ever hitting $a>x$, and identify which part of the transform formula gives it.
For zero-drift Brownian motion started at the midpoint of a bounded interval, what is the probability of exiting at the upper endpoint?
For $\mu\ne0$, verify that the displayed mean-exit formula is zero at both interval endpoints.
Why does boundedness of the stopped stochastic-integral integrand justify taking expectations in the finite-time Ito calculation?
In the optional-stopping example, which unjustified step leads to the contradiction $e^{-2}=1$?

Question	Quantity	Key condition or distinction
Ever hit an upper level?	\(\mathbb{P}(\tau_a<\infty)\)	Drift sign determines whether the probability is below one
Discounted first passage?	\(\mathbb{E}[e^{-q\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]\)	\(q\ge0\); at \(q=0\) this is the hitting probability
Which interval endpoint?	\(\mathbb{P}(X_\tau=u)\)	Use the driftless limit separately when \(\mu=0\)
How long until exit?	\(\mathbb{E}[\tau]\)	Finite for a bounded interval with \(\sigma>0\)